09-214 Physical Chemistry Name_______KEY___________________

Quiz 7 March 16, 2001

1 atm = 760 torr

R = 8.314 J/K mol

= 0.0821 L atm/K mol

(6 pts) A solution of the artificial sweetener sucrosa (mol wt 201.2) is prepared by adding 20.12 g of sucrosa to 100.0 g of water (mol wt 18.02). The vapor pressure of pure liquid water is 17.54 torr at 20^oC. Sucrosa is nonvolatile. Assuming that aqueous sucrosa solutions behave ideally, calculate the vapor pressure lowering (the difference between the vapor pressure of pure water and that of the solution) for the above solution at 20^oC.

n_B = 20.12/201.2 = 0.1000 mol

n_A = 100.0/18.02 = 5.55 mol

x_B = 0.1000/(0.1000 + 5.55) = 0.0177

P_A = x_AP_A^o = (1 - x_B) P_A^o = P_A^o - x_BP_A^o

P_A^o - P_A = x_BP_A^o = 0.0177(17.54 torr) = 0.31 torr

(3 pts) Whoops! Somebody points out that sucrosa is actually a sodium salt that ionizes completely in water, as symbolized by NaSu Æ Na⁺(aq) + Su^-(aq). In what way, if any, does this change your answer?

Both Na⁺ and Su^- contribute to vapor pressure lowering: n_B = 2(0.1000) mol

x_B = 0.2000/(0.2000 + 5.55) = 0.0348

P_A^o - P_A = x_BP_A^o = 0.0348(17.54 torr) = 0.61 torr

(6 pts) Whoops again! That same pest points out that the temperature is actually 25^oC. If the molar heat of vaporization of water is 44.0 kJ/mol, how does the higher temperature affect your answer to the first part?

ln (P₂/P₁) = - ( H/R)[(1/T₂) - (1/T₁)] = - (44000 J/mol/ 8.314 J/K mol)[(1/298)-(1/293)] = +0.303

P₂/P₁ = e^+0.303 = 1.35

P₂ = 1.35 P₁ = 1.35 (17.54 torr) = 23.7 torr (vapor pressure of pure water at 25^oC)

P_A^o - P_A = x_BP_A^o = 0.0177(23.7 torr) = 0.42 torr